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field gun will hit

by sanju09 » Thu Feb 09, 2012 4:44 am
What is the probability that a field gun will hit thrice on target in five tryouts?
I. The field gun maintains a history of hitting once on target in every five tryouts.
II. The probability that the field gun will hit twice on target in five tryouts is 0.2048.





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Source: — Data Sufficiency |
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by MBACenter » Mon Feb 13, 2012 6:21 am
sanju09 wrote:What is the probability that a field gun will hit thrice on target in five tryouts?
I. The field gun maintains a history of hitting once on target in every five tryouts.
II. The probability that the field gun will hit twice on target in five tryouts is 0.2048.





[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]
Statement 1 is sufficient. Statement 1 tells us that the probability of hitting on target in any one shot is 1/5. If we want the probability that the gun hits the target three times and misses twice, this would be:

(1/5) * (1/5) * (1/5) * (4/5) * (4/5) * 5!/(3!2!) = 32/625

Why the last step, multiplying by 5!/(3!2!)? Because we need to account for all the different scenarios and orderings of wins and losses (as though writing a 5-letter word with W, W, W, L, L).

Statement 2 is sufficient. Let's call the probability of hitting on any given shot P: the probability of missing on any given shot becomes 1 - P. Using the logic above, the probability of winning on two shots and losing on the others (remember: W, W, L, L, L) would be:

(P) * (P) * (1 - P) * (1 - P) * (1 - P) * 5!/(2!3!) = 2048/10,000

We should be able to solve for P from there. The correct answer is D.
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by sanju09 » Mon Feb 13, 2012 11:55 pm
rustypolymath wrote:
sanju09 wrote:What is the probability that a field gun will hit thrice on target in five tryouts?
I. The field gun maintains a history of hitting once on target in every five tryouts.
II. The probability that the field gun will hit twice on target in five tryouts is 0.2048.





[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]
Statement 1 is sufficient. Statement 1 tells us that the probability of hitting on target in any one shot is 1/5. If we want the probability that the gun hits the target three times and misses twice, this would be:

(1/5) * (1/5) * (1/5) * (4/5) * (4/5) * 5!/(3!2!) = 32/625

Why the last step, multiplying by 5!/(3!2!)? Because we need to account for all the different scenarios and orderings of wins and losses (as though writing a 5-letter word with W, W, W, L, L).

Statement 2 is sufficient. Let's call the probability of hitting on any given shot P: the probability of missing on any given shot becomes 1 - P. Using the logic above, the probability of winning on two shots and losing on the others (remember: W, W, L, L, L) would be:

(P) * (P) * (1 - P) * (1 - P) * (1 - P) * 5!/(2!3!) = 2048/10,000

We should be able to solve for P from there. The correct answer is D.
Can we expect a unique p from the equation p^2 (1 - p) ^3 = 0.2048?
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by [email protected] » Sun Mar 18, 2012 10:25 pm
  • Statement 1 is sufficient. Statement 1 tells us that the probability of hitting on target in any one shot is 1/5. If we want the probability that the gun hits the target three times and misses twice, this would be:

    (1/5) * (1/5) * (1/5) * (4/5) * (4/5) * 5!/(3!2!) = 32/625

    Why the last step, multiplying by 5!/(3!2!)? Because we need to account for all the different scenarios and orderings of wins and losses (as though writing a 5-letter word with W, W, W, L, L).

    Statement 2 is sufficient. Let's call the probability of hitting on any given shot P: the probability of missing on any given shot becomes 1 - P. Using the logic above, the probability of winning on two shots and losing on the others (remember: W, W, L, L, L) would be:

    (P) * (P) * (1 - P) * (1 - P) * (1 - P) * 5!/(2!3!) = 2048/10,000

    We should be able to solve for P from there. The correct answer is D.
In the statement 1. 5C2 is used and in the statement 2 5C3 is used. They should have been the same right...

Could you please help me in understanding this step...

5!/(3!2!) in statement 1 and 5!/(2!3!) in statement 2...

I have got the logic for it but not understanding the difference between them...
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by [email protected] » Sun Mar 18, 2012 10:33 pm
I think I got it...

In statement 1 we are focusing on 3 hits and 2 not hitting the target. Hence 5C3

In statement 2 we are focusing on 2 hits and 3 not hitting the target. Hence 5C2\



We are applying this formula as we also do not know which shot is going to hit out of the 5 tryouts.

I hope I am correct on this...
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by Anurag@Gurome » Sun Mar 18, 2012 11:28 pm
sanju09 wrote:What is the probability that a field gun will hit thrice on target in five tryouts?
I. The field gun maintains a history of hitting once on target in every five tryouts.
II. The probability that the field gun will hit twice on target in five tryouts is 0.2048.

[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]
Suppose a binomial experiment consists of n trials and results in x successes. If the probability of success on an individual trial is P, then the binomial probability is = nCx * P^x * (1 - P)^(n - x)

Here we have to find the value of 5C3 * P^3 * (1 - P)^(5 - 3) = 5C3 * P^3 * (1 - P)².
Now we need to know the value of P.

(1) The field gun maintains a history of hitting once on target in every five tryouts implies P = 1/5; SUFFICIENT.

(2) The probability that the field gun will hit twice on target in five tryouts is 0.2048 implies 5C2 * P² * (1 - P)^3 = 0.2048, solving which we'll get 3 values of P; NOT sufficient.

The correct answer is A.

I would like to mention here that the wording of this question is not like an actual GMAT question.
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by Mike@Magoosh » Mon Mar 19, 2012 8:06 am
I am posting in response to [email protected]'s request.

Prompt:
What is the probability that a field gun will hit thrice on target in five tryouts?

Statement #1: The field gun maintains a history of hitting once on target in every five tryouts

So, p = 1/5. Therefore, we could solve for three successes in five trials --- there's no reason to perform the actual calculation, as this is Data Sufficiency. It's enough to know that we could perform the calculation. Statement #1 is sufficient, as I believe everyone in this post agrees.

Statement #2: The probability that the field gun will hit twice on target in five tryouts is 0.2048

First of all, I am going to say: this question is way out-of-bounds in terms of difficulty. Even given a calculator, this would be a tricky question, but without a calculator, it's fiendish.

P(X = 2) = 0.2048 = (2^11)/10000 = (2^7)/625 = 128/625
P(X = 2) = (5C3)(p^2)(1-p)^3 = 128/625
10(p^2)(1-p)^3 = 128/625
That's a quintic equation, far more difficult that anyone short of Kepler would be expected to solve without a calculator.

With a calculator, we can graph the function f(x) = 10(x^2)(1-x^3, and see that it has two solutions for f(x) = 0.2048 for 0 < x < 1 ---- those two are x = 0.2 and x = 0.6262067027. So, it has more than one root in the realm that meaningful for probabilities. So, statement #2 is insufficient.

Answer = A

Again, it is absolutely beyond me how one would be supposed to solve the equation in #2 without a calculator. Yes, it is a quintic, which guarantees five solutions in the complex plane, but just from casual inspection, we don't know how many solutions are real, and we don't know how many of the real solutions are between zero and one.

Let me know if anyone has any questions on what I've said here.

Mike :)
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